Solution

The shaft ( radius R) spins with angular velocity $\omega$ , but the sleeve (bushing) is prevented from rotating by the counterweight. So at the contact surface between shaft and sleeve there is relative sliding made of two perpendicular components:

1. a circumferential component, from the shaft's rotation: $u = \omega R$
2. an axial component, from the sleeve's motion along the shaft: $v$ ( the steady-state velocity we want )

Since these two velocity components are mutually perpendicular ( one is along the surface's circumference, the othe along the axis), the resultant relative sliding speed is:
$$u_{rel} = \sqrt{ v^2 + ( \omega R)^2 }$$

Kinetic friction always acts opposite to the relative sliding velocity at the contact, and its magnitutde equals the maximum friction force $F_{tr}$ (given). So the friction force vector has magnitude $F_{tr}$, directed opposite to $\vec{u}_{rel}$ .

The axial component of this friction force (the part that resists the applied force $F$) is the projection of $F_{tr}$ along the axis:

$$F_{tr, axial} = F_{tr} \ \cdot \ \frac{v}{u_{rel}} = F_{tr} \ \cdot \ \frac{v}{\sqrt{ v^2 + (\omega R)^2}}$$
(this follows just from similar triangles: the axial component of friction is to $F_{tr}$ as $v$ to $u_{rel}$)

"Steady-state" means the bushing moves with constant velocity, so the net exial force is zero: the applied force $F$ is exactly balanced by the avial component of friction:

$$F = F_{tr} \ \cdot \ \frac{v}{\sqrt{ v^2 + (\omega R)^2}}$$

Now solve for $v$:

$$F \sqrt{v^2 + \omega^2 R^2} = F_{tr} \ v$$

Square both sides:

$$F^2 (v^2 + \omega^2 R^2 ) = F_{tr}^2 v^2$$
$$F^2 \omega^2 R^2 = v^2 (F_{tr}^2 - F^2)$$